Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 189

Alkynes are less reactive than alkenes towards electrophilic reagents. But the alkyne chemistry is useful for organic synthesis, in the form of hydration of the triple bond, formation of metalacetylides, selective reduction of alkynes and few other related reactions. Acetylene and HCHO react together in the presence of \(\mathrm{NaOCH}_{3}\) to give (a) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{OH}\) (b) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH}\) (c) \(\mathrm{HOH}_{2} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3}-\mathrm{CHOH}-\mathrm{CHOH}-\mathrm{CH}_{3}\)

Short Answer

Expert verified
Answer: (b) H2C=CH-CH2OH

Step by step solution

01

Write the reaction equation

Write down the reaction between acetylene and HCHO in the presence of NaOCH3. Acetylene has the formula C2H2, and HCHO is formaldehyde (CH2O). C2H2 + CH2O --[NaOCH3]--> Product
02

Recognize the nature of the reactants

The given reactants are acetylene, which is an alkyne, and formaldehyde, which is an aldehyde. Both of these are electrophilic in nature; however, the methoxide ion (OCH3-) from NaOCH3 acts as a nucleophile.
03

Analyze the possible reaction mechanism

The methoxide ion, being a nucleophile, attacks the electrophilic carbon atom of the formaldehyde and pushes the electrons towards the oxygen atom. This creates an intermediate where the methoxide ion is now bonded to the carbon atom, forming a methoxy alcohol group.
04

Attach the new functional group to the alkyne

The intermediate then reacts with acetylene, where the nucleophilic methoxy alcohol group attacks the electrophilic alkyne carbon atom, resulting in the addition of the new functional group across the triple bond. This forms a vinyl alcohol (enol) structure.
05

Identify the final product

Based on the above reaction mechanism, we find that the enol formed in the reaction is: H2C=CH-CH2OH Comparing this with the given options, we see that this product matches option (b). Thus, the correct answer is (b) H2C=CH-CH2OH.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The total number of chlorinated products given by ethane is (a) 6 (b) 7 (c) 8 (d) 9

Question contains Statement-1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct. (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement-1 (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1 (c) Statement- 1 is True, Statement- 2 is False (d) Statement-1 is False, Statement- 2 is True Statement 1 Alkynes are more reactive than alkenes in electrophilic addition reactions. and Statement 2 Electrophilic addition to both alkenes and alkynes involve a three membered cyclic intermediate and the bromonium ion type intermediate is less stable for an alkyne than for an alkene.

When \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{C}-\mathrm{CH}_{3}\) is treated with ozone by reductive ozonolysis procedure the product(s) obtained is/are (a) \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{COCO} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CHO}\) (d) \(\mathrm{CH}_{3} \mathrm{COOH}\)

Alkanes and alkenes can be prepared by (a) Wurtz reaction (b) Williamson synthesis (c) Dehydrohalogenation (d) Kolbe's electrolysis

Identify the correct statement among the following. (a) A hydrocarbon of M.F. \(\mathrm{C}_{6} \mathrm{H}_{12}\) on reductive ozonolysis gave only one product which fails to react with Tollens reagent. The hydrocarbon could be 2,3 dimethyl but- 2 -ene. (b) Addition of bromine to trans-2-butene gives a racemic mixture of two optically active enantiomers. (c) But-1-yne and but-2-yne can be distinguished using ammoniacal cuprous chloride. (d) Ozonolysis of acetylene gives glyoxal.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free